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1Z0-811 Arrays and Methods Practice Question

Exhibit

javac Test.java
Test.java:5: error: reference to print is ambiguous
        print(null);
        ^
  both method print(String[]) in Test and method print(String...) in Test match

Refer to the exhibit. Which overloaded methods cause this compilation error?

⚠ Common exam trap

The trap here is that candidates mistakenly think varargs and arrays are distinct types for overloading, but Java treats them identically after compilation, so defining both `print(String[])` and `print(String...)` causes a duplicate method error.

Answer choices

Why each option matters

Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.

Correct answer & explanation

✓

public void print(String[] s) and public void print(String... s)

Java does not allow overloading methods that differ only by the use of a varargs parameter and an array parameter of the same type. Both `public void print(String[] s)` and `public void print(String... s)` have the same method signature after erasure — they both accept a `String[]` at the bytecode level — causing a compilation error due to ambiguity.

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

  • ✗

    public void print(String s) and public void print(String[] s)

    Why it's wrong here

    This combination is valid because the first method takes a single String and the second takes a String array; they have different parameter types, so no compilation error.

  • ✗

    public void print(String s) and public void print(String... s)

    Why it's wrong here

    This combination is valid because a single String parameter and a varargs String parameter are different signatures; the compiler can distinguish between them, so no error.

  • ✓

    public void print(String[] s) and public void print(String... s)

    Why this is correct

    This combination causes a compilation error because a String array parameter and a String varargs parameter are considered the same after type erasure; the compiler sees them as duplicate methods.

  • ✗

    public void print(Object... o) and public void print(String... s)

    Why it's wrong here

    This combination is valid because the varargs are of different types (Object and String), so the method signatures are distinct; no compilation error.

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Senior Network & Security Engineer · founder of Courseiva

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